3
\$\begingroup\$

I have seen two parallel resistors used in series with LEDs in several LED bulbs. Why are parallel resistors used instead of series one?

enter image description here

\$\endgroup\$
  • 13
    \$\begingroup\$ That is, by the way, a bad circuit for powering LEDs. If you have LEDs in parallel, then each branch should have its own current limiting resistor. The forward voltage of each LED is a little different. So, one will light up before the others and snarf up all the current. That can cause it to burn out. Then you have fewer LEDs for the available current so it becomes more likely that another will burn out. So it keeps geeting worse until all the LEDs are burned out. \$\endgroup\$ – JRE Jun 24 '18 at 9:26
  • 4
    \$\begingroup\$ Please don’t parallel LEDs. You’re going to have a bad time. \$\endgroup\$ – winny Jun 24 '18 at 10:13
  • 2
    \$\begingroup\$ JRE, winny, Uwe - Paralleling LEDs in commercial "light engines" is common practice. They're all guaranteed to be from the same production batch and sharing the same 'heatsink' (PCB or whatever). \$\endgroup\$ – brhans Jun 24 '18 at 23:47
  • 2
    \$\begingroup\$ @brhans It's a common practice, indeed. That doesn't mean that a) it's a good practice, b) a non-commercial user should ever go that route. \$\endgroup\$ – user20088 Jun 25 '18 at 0:25
  • 3
    \$\begingroup\$ @Misunderstood: I have a drawer full of LED flashlights who argue otherwise. I agree with brhans that it can work with matched LEDs, but that's a rare case. If you build something at home, then you will find it difficult to get matched LEDs - separate current limiting is needed. And, judging by the number of dead flashlights I've accumulated, the manufacturers of cheap flashlights either can't get matched LEDs, or don't bother. \$\endgroup\$ – JRE Jun 26 '18 at 5:39
16
\$\begingroup\$

Because doubling the current paths halves the power dissipation in each path, allowing use of cheaper resistors.

| improve this answer | |
\$\endgroup\$
  • 5
    \$\begingroup\$ But paralleling resistors does not solve the problems of paralleled LEDs. \$\endgroup\$ – Uwe Jun 24 '18 at 16:11
  • \$\begingroup\$ The same is true for resistors-in-series.... \$\endgroup\$ – ThreePhaseEel Jun 25 '18 at 1:39
  • \$\begingroup\$ @Uwe and others -- the question is NOT about parallel LEDs, just about parallel resistors. (Although I agree about each LED having its own resistor.) \$\endgroup\$ – Jennifer Jun 25 '18 at 5:26
  • 1
    \$\begingroup\$ @Uwe except there most likely is NO problem with parallel LEDs. \$\endgroup\$ – Misunderstood Jun 26 '18 at 4:30
10
\$\begingroup\$

You mean instead of single resistor? It depends on the use case.

Assume a power of 600 mW has to be dissipated across the resistor. Instead of choosing one single resistor of standard wattage, say 1 W, one will choose a resistor (double value of intended resistance) and a power wattage of 500 mW..

Effectively, you have same resistance value and you are distributing power dissipation in two resistors instead of one.. These two resistors price combined may be lesser than one single bulky resistor because 500 mW resistors may be used widely compared to 1W resistors.

Also, you have now chance to use multiple resistance values.. Parallel combination of multiple resistors can lead to same value of required resistance. You can re use the parts used in other designs, also manage if the currently used resistor is no more available.

It also distributes the heat to wider region than compared to single resistor avoiding local heating.


If LEDs are in parallel, and assuming random LEDs which is more probable for a lab user:

This is how it will fail sooner or later:

  1. User will power on the board
  2. The LEDs will need certain voltage (forward voltage, Vf ) to turn on.
  3. Assume all LEDs have VD between 1.3 and 1.6
  4. The LED with least Vf (1.3 V) will turn on first
  5. This keeps the voltage across all remaining LEDs at 1.3 V
  6. It means entire current is flowing through only one LED!! As other LEDs are off (or dim)
  7. This single LED will be very bright, also raises its temperate due to heat dissipation
  8. Raise in temperature leads to further drop in its forward voltage.
  9. Further drop in forward voltage leads to further increase in current (also, remaining LEDs will be completely off)
  10. LED burns out eventually
  11. Soon after that the LED with next low forward voltage (say 1.35V) will turn on step 5 to 11 will repeat for this LED and all other LEDs in parallel
  12. This may happen instantly or over a period of time

Adding individual resistors in series with each LEDs helps in bringing down this sequence by limiting the current.

Why not connect? Useful answer here

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ You were fine up to the last paragraph. Paralleling resistors doesn't change how current gets to individual paralleled LEDs. \$\endgroup\$ – Graham Jun 25 '18 at 7:46
  • \$\begingroup\$ paralleling and serial connection terminology is confusing in this whole discussion. removed it \$\endgroup\$ – User323693 Jun 25 '18 at 9:59
  • \$\begingroup\$ Agreed - you've cleared that up. \$\endgroup\$ – Graham Jun 25 '18 at 12:59
  • 2
    \$\begingroup\$ It is true the LED with the lowest individual forward voltage will draw the most current. But not all. Point #2 is false, LEDs do not need a certain voltage to turn on. In this case all LEDs will be on. The amount of current flowing determines the Vf in a wide range of the IV curve. Your point #5, #6, #9 (completely off) are false. Your whole scenario is wrong based on false statements. #11 and #12 will never happen because #6 is false. You also assume the Vf of the LEDs are significantly different. Most lighting LEDs can be purchased where Vf is binned. 1.3v is a strange example Vf. \$\endgroup\$ – Misunderstood Jun 25 '18 at 23:13
  • 1
    \$\begingroup\$ Downvote for the FUD, as noted by misunderstood. \$\endgroup\$ – Passerby Jun 26 '18 at 4:35
0
\$\begingroup\$

Most likely reason is inventory.
I have 1000's of 5.1Ω 5W resistors so I would use two of them if the current was an amp or more.

Maybe two at half value is cheaper than one, likely not true especially if Vcc is 3.3V. Not easier or cheaper to assemble.

To make any other comment on the validity of the circuit is fruitless due to the unknowns such as Vcc and LED part number.

Without an known value for Vcc most of the comments made so far are invalid. What is the difference if Vcc were 3.3V vs. 12V?

Parallel current mismatch is more of an issue with strings of multiple high power LEDs wired in parallel. Single mid-power LEDs wired in parallel today rarely creates a problem.

Today connecting single LEDs is parallel is very common and works very well. Samsung and Bridgelux sell many high quality strips with parallel LEDs and they have no problem. A few years ago it was a different story.

LEDs today are producing 80% light, 20% heat, with very little variation in Vf, so many of the comments made are no longer valid.

Then there is this "6. It means entire current is flowing through only one LED!! As other LEDs are off (or dim)". So NOT true.

The LEDs have different Vf when measured individually. When wired in parallel they all are operating at the same Vf. The combined Vf is that of the highest individual Vf. It is true the LED with the lowest individual Vf will be drawing more current, but all will be on and the difference in perceived intensity will likely be indistinguishable.

For example Samsung's best selling F-Series Gen3 strips are wiring sets of 9 or 18 LEDs in parallel.

The issue is not LEDs wired in parallel it's about failure recovery: OSRAM App note LED circuit comparison

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ "LEDs today are producing 80% light, 20% heat, with very little variation in Vf, so many of the comments made are no longer valid" Please show with an example. Hint: They don't exist. Highest I've seen is 55% for a blue emitter. \$\endgroup\$ – winny Jun 26 '18 at 7:22
  • \$\begingroup\$ @winny Cree’s new XLamp® XP-G3 Royal Blue LED doubles the maximum light output of similar size competing LEDs and delivers breakthrough wall-plug efficiency of up to 81%. led-professional.com/products/leds_led_modules/… \$\endgroup\$ – Misunderstood Jun 26 '18 at 13:50
  • \$\begingroup\$ @winny OSRAM GD CSHPM1.14, Vf 2.75v, 765. mW @ 350 mA = 79.5% efficacy (69% typical). dammedia.osram.info/media/resource/hires/osram-dam-4486463/… These two examples are high power LEDs used in white pushing about 200 lm/W. Mid-power are currently the more efficient, e.g. Samsung LM301B, 218 lm/W typ. @ 65 mA. Available with a Vf range from 2.6Vmin-2.7Vmax over the full operating range of 10-200 mA. \$\endgroup\$ – Misunderstood Jun 26 '18 at 14:26
  • \$\begingroup\$ Neat. Still a far cry from that today's led are anywhere near 80% light. We are talking about white LEDs here. \$\endgroup\$ – winny Jun 26 '18 at 14:48
  • \$\begingroup\$ White LEDs use deep/royal blue LEDs with a yellow phosphor. Therefore white LEDs emit the same number of photons from the band gap as deep blue becasue they are deep blue. And as I said the LM301B (mid-power white) has better efficacy than the XP-G3. "We are talking about white LEDs here" is an assumption on your part. The LEDs here are drawn in blue ink. I have measured lux, radiometric watts, and µmols from white LEDs 3000K to 5700K and all emit about the same number of photons (quantum 60 µmol/s/m²) and watts (13-14 radiometric watts) while lux (efficacy) ranged 5100-5625 lux. \$\endgroup\$ – Misunderstood Jun 26 '18 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.